Explore BrainMass
Share

Elementary Statistics for Null and Alternative Hypothesis

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Elementary Statistics

1. A researcher knows that the population mean among college students taking the social test is a "neutral" 100. (scores higher than 100are assumed to represent more conformity than average and those lower than 100 indicates less conformity). A random sample of 30 students was selected, and the following values recorded:
Sample mean=103
Sample standard deviation=10.83
Is this sample significantly different than the population?
State the null and alternative hypothesis.

2. A recent national survey reports that the general population gives the president an average rating of mean=62 on a scale of 1 to 100. A researcher suspects that college students are likely to be more critical of the president than people in the general population. To test these suspicions, a random sample of college students is selected and asked to rate the president. The data for this sample are as follows:
44, 52, 24, 45, 39, 57, 20, 38, 78, 74, 61, 56, 49, 66, 53, 49, 47, 88, 38, 51, 65, 47, 35, 59, 23, 41, 50, 19.
On the basis of this sample, can the researcher conclude that college students rate the president differently? Test at the .01 level of significance, two tailed.

3. A family therapist states that father talk to their teenagers on average of twenty seven minutes per week. Surprised by that claim, a psychologist decided to collect some data on the amount of time fathers spend in conversation with their children. For n=12, parents the study revealed the following times (in minutes) devoted to conversation in a week: 29, 22, 19, 25, 27, 28, 21, 22, 24, 26, 30, 22.
Do the psychologist's findings differ significantly from the therapist's claim? If so, is the family expert's claim an overestimate or underestimate of actual time spent talking to children? Use the.05 level of significance, two tailed.

4. For the following data, calculate the Pearson Correlation.
Person X Y
A 0 4
B 2 1
C 8 10
D 6 9
E 4 6

5. Do weather conditions affect peoples' moods? To examine this question, a researcher selected three samples of college students and administered a mood inventory questionnaire to each student. One group was tested on a dreary, overcast and drizzling day. The second group was tested during a violent thunderstorm and the third group was tested during a violent thunderstorm, and the third group was tested on a bright sunny day. The data is as follows:
Dreary: 6, 9, 10, 12, 5, 7, 12, 8, 7, 10
Stormy: 8, 12, 10, 6, 8, 9, 14, 10, 7, 7
Bright: 13, 10, 6, 13, 10, 8, 9, 12, 15, 11
Does the data indicate that weather has an effect on mood? Test at the .05 level of significance

6. A researcher wanted to know if there is a relationship between arousal and performance. The general result of similar studies shows that increasing the level of arousal tends to improve the level of performance. The data given has three levels of arousal and two levels of task difficulty. Does this data indicate a significant relationship between arousal and performance? Does the level of difficulty affect performance?
AROUSAL LEVEL
Low Medium High
3 2 9
1 5 9
EASY 1 9 13
6 7 6
4 7 8
0 3 0
DIFFICULT 2 8 0
0 3 0
0 3 5
3 3 0

7. A random sample of 100 adult workers 60 men and 40 women are selected in a large city. Of the group 42 men and 17 women are union members. Does the union membership differ on the basis of gender? State the independent and dependant variables.

8. A random sample of ninety college students indicates whether they most desire love, wealth, power, health, fame or family happiness. Using the .05 level of significance test the null hypothesis in the underlying population the various desires are equally popular.
Desires of College Students
Love: 25 Wealth: 10
Power: 5 Heath: 25
Fame: 10 Family Happiness: 15

9. A social scientist cross-classifies the responses of 100 randomly selected people on the basis of gender and whether or not they favor strong gun control laws. She obtained the following data:
Favor Oppose
Male 40 20
Female 30 10
Test the null hypothesis for a relationship between gender and attitude toward gun control.

10. The correlation between IQ and the Verbal SAT Score among the seniors at GWB High School is .90; it is significant. The mean IQ is 100 with a standard deviation of 100. Predict SAT Scores for Seniors with the following IQ's: 105, 110, 120

© BrainMass Inc. brainmass.com October 10, 2019, 12:17 am ad1c9bdddf
https://brainmass.com/statistics/hypothesis-testing/elementary-statistics-null-alternative-hypothesis-285940

Solution Preview

Please see the attached files

Elementary Statistics

Given the population mean  = 100; sample mean = 103; sample standard deviation s = 10.83 and sample size n = 30

The hypothesis to be tested is against
The test statistic is given by . This statistic follows student's t distribution with n-1 degrees of freedom.

The calculated value of the statistic is =1.5172
The p value from student's t tables with 29 degrees of freedom is 0.1400. Since the p value is greater than 0.05 the calculated value of t is not significant. Hence we accept the null hypothesis.

Note: Alternatively the critical value from the student's t tables with 29 degrees of freedom is 2.0452. Since the modulus of the calculated value is less than critical value we accept the null hypothesis.

Conclusion: Sample is not significantly different from the population

Here the average rating of the population is given as 62 i.e., = 62. A sample of size 28 is given. It can be calculated that sample mean = 48.8571; sample standard deviation s = 16.0372

The hypothesis to be tested is against at  = 0.01
The test statistic is given by . This statistic follows student's t distribution with n-1 degrees of freedom.
The calculated value of the statistic is = - 4.34
The p value from student's t tables with 27 degrees of freedom is 0.0002 . Since the p value is less than 0.01 the calculated value of t is significant at 1% level. Hence we reject the null hypothesis.

Note: Alternatively the critical value from the student's t tables with 27 degrees of freedom is 2.77when  =0.01. Since the modulus of the calculated value is greater than critical value reject the null hypothesis.

Conclusion: The data support the argument that college students rate the president differently.

3. Here the mean talking time of the population is given as 27. i.e., = 27; A sample of size 12 is given. It can be calculated that sample mean = 24.58; sample standard deviation s = 3.48

The hypothesis to be tested is against at  = 0.05
The test statistic is given by . This statistic follows student's t distribution with n-1 degrees of freedom.

The calculated value of the statistic is = -2.41

The p value from student's t tables with 11 degrees of freedom is 0.0347. Since the p value is less than 0.05 the calculated value of t is significant at  = 0.05. Hence we reject the ...

Solution Summary

The expert examines elementary statistics for Null and Alternative hypothesis.

\$2.19